Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{\mathbf{1}}{{\left(s^2+4\right)}^{\mathbf{2}}}$

Let $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and $\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}$be piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and $\mathit{\alpha }$of exponential order and set $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{f}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{g}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$, then,

or

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\mathbf{\left\{}\mathit{F}\mathbf{\right(}\mathit{s}\mathbf{\left)}\mathit{G}\mathbf{\right(}\mathit{s}\mathbf{\left)}\mathbf{\right\}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathit{f}\mathbf{*}\mathit{g}\mathbf{)}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

Consider the given function,

$\frac{1}{{\left(s^2+4\right)}^2}$

Let,

$$\begin{gathered} F\left(s\right)=G\left(s\right) \\ =\frac{1}{s^2+4} \end{gathered}$$

Hence, we know,

$$\begin{gathered} f\left(t\right)=g\left(t\right) \\ ={\mathcal{L}}^{-1}\left\{\frac{1}{s^2+4}\right\}\left(t\right) \\ =\frac{\sin 2t}{2} \end{gathered}$$

Take inverse Laplace and use the convolution theorem,

$$\begin{gathered} \left.{\mathcal{L}}^{-1}\left\{\frac{1}{{\left(s^2+4\right)}^2}\right\}\left(t\right)={\mathcal{L}}^{-1}F\left(s\right)G\left(s\right)\right\}\left(t\right) \\ =f\left(t\right)*g\left(t\right) \\ =\frac{\sin 2t}{2}*\frac{\sin 2t}{2} \end{gathered}$$

Apply the convolution formula, $(f*g)\left(t\right)={\int }_0^{t}f(t-v)g\left(v\right)dv$

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{1}{{\left(s^2+4\right)}^2}\right\}\left(t\right)={\int }_0^t\frac{{\displaystyle \sin 2(t-v)}}{{\displaystyle 2}}\frac{{\displaystyle \sin 2v}}{{\displaystyle 2}}dv \\ =\frac{1}{4}{\int }_2^t\sin 2(t-v)\sin 2vdv \\ =\frac{1}{8}{\int }_0^t\left(\cos \right(2t-4v)-\cos 2t)dv \\ =\frac{1}{8}{\int }_0^t\cos (2t-4v)dv-\frac{1}{8}\cos 2t \end{gathered}$$

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{1}{{\left(s^2+4\right)}^2}\right\}\left(t\right)=\frac{1}{8}{\left[\frac{\sin (4v-2t)}{4}\right]}_0^t-\frac{\cos 2t}{8}[-v{]}_0^t \\ =\frac{\sin 2t}{16}-\frac{t\cos 2t}{8} \end{gathered}$$

Therefore, the inverse Laplace transform for the given function is

${\mathcal{L}}^{-1}\left\{\frac{1}{{\left(s^2+4\right)}^2}\right\}\left(t\right)=\frac{\sin 2t}{16}-\frac{t\cos 2t}{8}$